Question

A geostationary satellite is orbiting the Earth at a height of $6R$ above the surface of the earth, $R$ being the radius of the Earth. The time period of another satellite, at a height of $\text{2.5}R$ from the surface of the Earth, is

• A. $6\sqrt{2} \, \text{hour}$
• B. $6 \, \text{hour}$
• C. $5\sqrt{2}\text{hour}$
• D. $10 \, \text{hour}$

Answer 1

Answer: (A)

A geostationary satellite has a time period of $T=24 \, hr$ . From Kepler's law,
$T^{2} \propto \, R^{3} \, \, \Rightarrow \frac{T_{2}^{2}}{T_{1}^{2}}=\frac{R_{2}^{3}}{R_{1}^{3}} \, \, \Rightarrow \frac{T_{2}^{2}}{\left(24\right)^{2}}=\frac{\left(R + \text{2.5} R\right)^{3}}{\left(R + 6 R\right)^{3}}$
$T_{2}=\sqrt{\frac{24 \times 24}{8}}=6\sqrt{2} \, hr$ .